Respuesta :

naǫ
A geometric series:
[tex]\sum^{\infty}_{i=1}=a_1 \times r^{i-1}[/tex]
It's convergent if |r|<1.
It's divergent if |r|≥1.
The sum can be found if it's a convergent series; it's equal to [tex]\frac{a_1}{1-r}[/tex].

3.
[tex]\sum^{\infty}_{i=1} 12 (\frac{3}{5})^{i-1} \\ \\ a_1=12 \\ r=\frac{3}{5} \\ \\ |r|<1 \hbox{ so it's convergent} \\ \\ \sum^{\infty}_{i=1} 12 (\frac{3}{5})^{i-1}=\frac{12}{1-\frac{3}{5}}=\frac{12}{\frac{5}{5}-\frac{3}{5}}=\frac{12}{\frac{2}{5}}=12 \times \frac{5}{2}=6 \times 5=30[/tex]

The answer is: This is a convergent geometric series. The sum is 30.

4.
[tex]\sum^{\infty}_{i=1} 15(4)^{i-1} \\ \\ a_1=15 \\ r=4 \\ \\ |r| \geq 1 \hbox{ so it's divergent}[/tex]

The answer is: This is a divergent geometric series. The sum cannot be found.

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